Disjoint unions and Cartesian products of combinatorial classes

Let $\mathcal{A}$ and $\mathcal{B}$ be combinatorial classes.

Their disjoint union $\mathcal{A}+\mathcal{B}$ is the tagged union

$\mathcal{A}+\mathcal{B}:=(\{0\}\times \mathcal{A})\cup (\{1\}\times \mathcal{B}),$

with size $\mathrm{\mid }(0,a)\mathrm{\mid }:=\mathrm{\mid }a\mathrm{\mid }$ and $\mathrm{\mid }(1,b)\mathrm{\mid }:=\mathrm{\mid }b\mathrm{\mid }$. The tags are part of the data: they keep the two copies disjoint even when $\mathcal{A}$ and $\mathcal{B}$ have common underlying objects.

Their Cartesian product $\mathcal{A}\times \mathcal{B}$ is the set of ordered pairs $(a,b)$ with $a\in \mathcal{A}$ and $b\in \mathcal{B}$, equipped with the size map

$\mathrm{\mid }(a,b)\mathrm{\mid }:=\mathrm{\mid }a\mathrm{\mid }+\mathrm{\mid }b\mathrm{\mid }\mathrm{.}$

Here the ordered pair itself records both components. This uniqueness of the factorisation is part of the construction: later counterexamples show that dropping it breaks the product rule.

For later shorthand, $m\mathcal{A}$ means the disjoint union of $m$ tagged copies of $\mathcal{A}$.